题目如图,∠ABC=∠ACB\angle ABC=\angle ACB∠ABC=∠ACB,ADADAD、BDBDBD、CDCDCD分别平分△ABC\triangle ABC△ABC的外角∠EAC\angle EAC∠EAC、内角∠ABC\angle ABC∠ABC、外角∠ACF\angle ACF∠ACF.以下结论:①ADADAD∥BC;BC;BC;②∠ACB=2∠ADB\angle ACB=2\angle ADB∠ACB=2∠ADB;③∠ADC=90∘−∠ABD\angle ADC=90^{\circ}-\angle ABD∠ADC=90∘−∠ABD;④∠BDC=∠BAC\angle BDC=\angle BAC∠BDC=∠BAC.其中正确的结论有( )A.111个B.222个C.333个D.444个知识点:平行线的性质章节:图形的性质 / 相交线与平行线